Question

Solve it integral 0 to pi x/1-cos alpha sin x)dx

Answer 1

I = $\frac{\pi}{sin\alpha}(\pi-\alpha)$

we have to solve integral $\int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx$

first apply application : when $\int\limits^a_b{f(x)}\,dx$

then, f(x) = f(a + b - x)

so, I = $\int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx=\int\limits^{\pi}_0{\frac{\pi-x}{1-cos\alpha sin(\pi-x)}}\,dx$

so, 2I = $\int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx+\int\limits^{\pi}_0{\frac{\pi-x}{1-cos\alpha sinx}}\,dx$

⇒2I = $\int\limits^{\pi}_0{\frac{\pi}{1-cos\alpha sinx}}\,dx$

now putting, sinx = 2tan(x/2)/(1 + tan²(x/2))

so, 2I = $\pi\int\limits^{\pi}_0{\frac{1}{1-cos\alpha\frac{2tan(x/2)}{1+tan^2(x/2)}}}\,dx$

⇒2I/π = $\int\limits^{\pi}_0{\frac{sec^2(x/2)}{1+tan^2(x/2)-cos\alpha 2tan(x/2)}}\,dx$

[ because, 1 + tan²(x/2) = sec²(x/2)]

now putting, tan(x/2) = t

so, sec²(x/2)(1/2) dx = dt

⇒sec²(x/2) dx = 2dt

upper limit : tan(π) → ∞

lower limits : tan(0) → 0

⇒2I/π = $\int\limits^{\infty}_0{\frac{2dt}{1+t^2-2cos\alpha t}}$

1 + t² - 2cosα t = (t - cosα)² - (cos²α- 1)

= (t - cosα)² + sin²α

so, I/π = $\int\limits^{\infty}_0{\frac{dt}{(t-cos\alpha)^2+sin^2\alpha}}$

= $\frac{1}{sin\alpha}\left[tan^{-1}\frac{1-cos\alpha}{sin\alpha}\right]^{\infty}_0$

so, I = $\frac{\pi}{sin\alpha}(\pi-\alpha)$

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